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3 years ago

Write two example in which chemical energy is converted into heat energy

Physics

1 answer:

natulia [17]3 years ago

5 0

Answer:

batteries in an electric heater, a nuclear power plant

Explanation:

sorry if it wrong g

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A(n) _______ is a point level measuring system consisting of a circuit of two or more probes or electrodes, or an electrode and

Digiron [165]

Answer:

Conductivity probe

Explanation:

The Conductivity Probe consists of two electrodes(also referred to as probes)or an electrode and a wall vessel where the material in the vessel completes the circuit as the level rises in the vessel.

It is used in measuring solution conductivity or total ionic concentration of aqueous samples.

6 0

3 years ago

Two isolated, concentric, conducting spherical shells have radii R1 = 0.500 m and R2 = 1.00 m, uniform charges q1=+2.00 µC and q

scZoUnD [109]

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

a E = $1.685*10^3 N/C$

b E = $36.69*10^3 N/C$

c E = 0 N/C

d $V = 6.7*10^3 V$

e $V = 26.79*10^3V$

f V = $34.67 *10^3 V$

g $V= 44.95*10^3 V$

h $V= 44.95*10^3 V$

i $V= 44.95*10^3 V$

Explanation:

From the question we are given that

The first charge $q_1 = 2.00 \mu C = 2.00*10^{-6} C$

The second charge $q_2 =1.00 \muC = 1.00*10^{-6}$

The first radius $R_1 = 0.500m$

The second radius $R_2 = 1.00m$

$Generally \ Electric \ field = \frac{1}{4\pi\epsilon_0}\frac{q_1+\ q_2}{r^2}$

And $Potential \ Difference = \frac{1}{4\pi \epsilon_0} [\frac{q_1 }{r}+\frac{q_2}{R_2} ]$

The objective is to obtain the the magnitude of electric for different cases

And the potential difference for other cases

Considering a

r = 4.00 m

$E = \frac{((2+1)*10^{-6})*8.99*10^9}{16}$

$= 1.685*10^3 N/C$

Considering b

$r = 0.700 m \ , R_2 > r > R_1$

This implies that the electric field would be

$E = \frac{1}{4\pi \epsilon_0}\frac{q_1}{r^2}$

This because it the electric filed of the charge which is below it in distance that it would feel

$E = 8*99*10^9 \frac{2*10^{-6}}{0.4900}$

= $36.69*10^3 N/C$

Considering c

r = 0.200 m

=> $r$

The electric field = 0

This is because the both charge are above it in terms of distance so it wont feel the effect of their electric field

Considering d

r = 4.00 m

=> $r > R_1 >r>R_2$

Now the potential difference is

$V =\frac{1}{4\pi \epsilon_0} \frac{q_1 + \ q_2}{r} = 8.99*10^9 * \frac{3*10^{-6}}{4} = 6.7*10^3 V$

This so because the distance between the charge we are considering is further than the two charges given

Considering e

r = 1.00 m $R_2 = r > R_1$

$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{1.00} \frac{1.00*10^{-6}}{1.00} ] = 26.79 *10^3 V$

Considering f

$r = 0.700 m \ , R_2 > r > R_1$

$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.700} \frac{1.0*10^{-6}}{1.00} ] = 34.67 *10^3 V$

Considering g

$r =0.500\m , R_1 >r =R_1$

$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V$

Considering h

$r =0.200\m , R_1 >R_1>r$

$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{R_1} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V$

Considering i

$r =0\ m \ , R_1 >R_1>r$

$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{R_1} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V$

8 0

3 years ago

A driven RLC circuit is being driven by an AC emf source with a maximum current of 2.75 A and maximum voltage of 150 V. The curr

weqwewe [10]

Answer:

(a). Z = 54.54 ohm

(b). R = 36 ohm

(c). The circuit will be Capacitive.

Explanation:

Given data

I = 2.75 A

Voltage = 150 V

$\phi = 0.85$ rad = 48.72°

(a). Impedance of the circuit is given by

$Z = \frac{V}{I}$

$Z = \frac{150}{2.75}$

Z = 54.54 ohm

(b). We know that resistance of the circuit is given by

$R = \frac{Z}{\sqrt{1 + \tan^{2}\phi } }$

Put the values of Z & $\phi$ in above formula we get

$R = \frac{54.54}{\sqrt{1 + \tan^{2} ( \ 48.72) } }$

R = 36 ohm

(c). Since the phase angle is negative so the circuit will be Capacitive.

3 0

3 years ago

A very long, uniformly charged cylinder has radius R and linear charge density λ. Find the cylinder's electric field strength ou

mixer [17]

The cylinder's electric field magnitude, at a distance <em>r</em> from the axis of the cylinder (greater than the cylinder's radius), is equal to $E= \frac{\lambda}{2\pi \epsilon_0 \cdot r}$

<h3>Further explanation</h3>

Matter is the building block of everything that we encounter in our lives. Matter is made of atoms, which are in turn made of tiny particles which are called electrons, protons, and neutrons. The ammount of these 3 elements, and their topological configuration in the atoms, is what determines what a certain element is (like Carbon, Hydrogen, Iron, etc).

In some cases, some elements may lose or gain some electrons. Regarded that this missing or extra electrons are not very high in number, the material doesn't lose any of its properties, however it will always try to get its number of electrons back to normal. This is when we say that an element has a <em>charge</em>, which is a measure of how much electrons a body needs to get back to normal. A body has positive charge if it lacks electrons, and has negative charge if it has extra electrons.

This charge causes the material to have an Electric field, which is a measure of how much does it attract or repel electrons. In the case of our problem, we need to compute exactly that, the Electric field. In our problem, we have an infinitely long cylinder with a linear charge density $\lambda$ , this means that all parts of the cylinder have the same charge, and due to symmetry, the electric field is constant on the angular and longitudinal directions of the cylinder.

This makes easy to apply Gauss' Law, since for a Gaussian curve in the shape of a concentric cylinder (with a higher radius than that of our charged cylinder) we can write:

$\Phi = \frac{\lambda \cdot L}{\epsilon_0}$

Where $\Phi$ is called the Electric flux. Since the electric field is constant for a given distance <em>r</em> from the axis of the cylinder we can write that:

$\Phi = E \cdot 2\pi r \cdot L$

Joining both our expressions we can get that:

$E= \frac{\lambda}{2\pi \epsilon_0 \cdot r}$

<h3 /><h3>Learn more</h3>

Description on Electric fields: brainly.com/question/8971780
Relation between electric fields and magnetism: brainly.com/question/2838625
How can we use electric charges: brainly.com/question/10427437

<h3>Keywords</h3>

Electrons, protons, electric field, cylinder, electric flux

5 0

3 years ago

Read 2 more answers

Which graph shows an object that is dropped?

Olegator [25]

Answer:

I'm pretty sure it's the third one where velocity goes from positive to negative

Explanation:

the positive velocity is before the object hits the ground and the negative is after

8 0

3 years ago

Write two example in which chemical energy is converted into heat energy ​

Write two example in which chemical energy is converted into heat energy